667 lines
34 KiB
TeX
667 lines
34 KiB
TeX
\section{Soundness}
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\newcommand{\CC}{\text{CC}}
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\begin{lemma}
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A sound TypelessFJ program is also sound under LetFJ type rules.
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\begin{description}
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\item[if:]
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$\Gamma | \Delta \vdash \texttt{m}(\ol{x}) = \texttt{e} \ \ok \ \text{in}\ C \text{with} \ \generics{\ol{Y \triangleleft P}}$
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\end{description}
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\end{lemma}
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TODO: Beforehand we have to show that $\Delta \cup \overline{\Delta} | \Theta \vdash \texttt{e} : \type{T} \mid \overline{\Delta}$
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Here $\Delta$ does not contain every $\overline{\Delta}$ ever created.
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%what prevents a free variable to emerge in \Delta.N example Y^Object |- C<String> <: X^Y.C<X>
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% if the Y is later needed for an equals: same(id(x), x2)
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Free wildcards do not move inwards. We can show that every new type is either well-formed and therefore does not contain any free variables.
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Or it is a generic method call: is it possible to use any free wildcards here?
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let empty
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<X> Box<X> empty()
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same(Box<?>, empty())
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let p1 : X.Box<X> = Box<?> in let
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X.Box<X> <. Box<x>
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Box<e> <. Box<x>
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boxin(empty()), Box2<?>
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Where can a problem arise? When we use free wildcards before they are freed.
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But we can always CC them first. Exception two types: X.Pair<X, y> and Y.Pair<x, Y>
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Here y = Y and x = X but
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<X,Y> void same(Pair<X,Y> a, Pair<X,Y> b){}
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<X> Pair<?, X> left() { return null; }
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<X> Pair<X, ?> right() { return null; }
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<X> Box<X> id(Box<? extends Box<X>> x)
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here it could be beneficial to use a free wildcard as the parameter X to have it later
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Box<?> x = ...
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same(id(x), id(x)) <- this will be accepted by TI
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let left : X,Y.Pair<X,Y> = left() in
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let right : Pair<X,Y> = right() in
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Compromise:
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- Generate constraints so that they comply with LetFJ
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- Propose a version which is close to Java
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Version for LetFJ:
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Is it still possible to do the capture conversion in form of constraints?
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X.C<X> <. C<x>
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T <. X.C<X>
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how to proof: X.C<X> ok
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If $\Delta \cup \overline{\Delta} | \Theta \vdash \texttt{e} : \type{T} \mid \overline{\Delta}$
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then there exists a $|\texttt{e}|$ with $\Delta | \Theta \vdash |\texttt{e}| : \wcNtype{\Delta'}{N}$ in LetFJ.
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This is possible by starting with the parameter types as the base case $\overline{\Delta} = \emptyset$.
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Each type $\wcNtype{\Delta'}{N}$ can only use wildcards already freed.
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\textit{Proof} by structural induction.
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\begin{description}
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\item[$\texttt{e} = \texttt{x}$] $\Delta | \Theta \vdash \texttt{e} : \type{T} \mid \emptyset$
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$\Delta \vdash \type{T} \ \ok$ by \rulename{T-Method}
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and therefore $\Delta | \Theta \vdash \texttt{let}\ \texttt{e} : \type{T} = \texttt{x in } \texttt{e}$.
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$|\texttt{x}, \texttt{e}| = \texttt{let}\ \texttt{e} : \type{T} = \texttt{x in } \texttt{e}$
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\item[$\texttt{e} = \texttt{e}.\texttt{m}(\ol{e})$] there must be atleast one value in $\texttt{e}$ or $\ol{e}$
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\item[$\texttt{e}.f$] given let x : T = e' in x
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let x : T = e' in let xf = x.f in xf
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Required:
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$ \Delta | \Theta \vdash e' : \type{T}_1$
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$\Delta \vdash \type{T}_1 <: \wcNtype{\Delta'}{N}$
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$\Delta, \Delta' | \Theta, x : \type{N} \vdash let xf = x.f in xf : \type{T}_2$
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\end{description}
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\textbf{Proof:} Every program complying with our type rules can be converted to a correct LetFJ program.
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First we convert the program so that every wildcards used in an expression are in the $\Delta$ environment:
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m(p) = e => let xp = p in [xp/p]e
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x1.m(x2) => let xm = x1.m(x2=) in xm
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x.f => let xf = x.f in xf
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Then we have to proof there is never a wildcard used before it is declared.
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Wildcards are introduced by the capture conversions and nothing else.
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\begin{theorem}\label{testenv-theorem}
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Type inference produces a correctly typed program.
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\begin{description}
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\item[If] $\fjtypeinference(\mv{\Pi}, \texttt{class}\ \exptype{C}{\ol{X}
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\triangleleft \ol{N}} \triangleleft \type{N}\ \{ \overline{\type{T} \ f};\ \ol{M} \}) = \mtypeEnvironment{}'$
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\item[Then] $\texttt{class}\ \exptype{C}{\ol{X}
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\triangleleft \ol{N}} \triangleleft \type{N}\ \{ \overline{\type{T} \ f};\ \ol{M} \} \text{ok}$,
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with $\ol{M} = $
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\end{description}
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\end{theorem}
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\begin{lemma}{Well-formedness:}
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TODO:
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\end{lemma}
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\begin{lemma}{Soundness:}
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\unify{}'s type solutions for a constraint set generated by $\typeExpr{}$ are correct.
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\begin{description}
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\item[if] $\typeExpr{}(\mtypeEnvironment{}, \texttt{e}, \tv{a}) = C$
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and $(\Delta_u, \sigma) = \unify{}(\Delta', C)$ and let $\Delta = \Delta_u \cup \Delta'$
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% , with $C = \set{ \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \type{S'} \lessdotCC \type{T'} } }$
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% and $\vdash \ol{L} : \mtypeEnvironment{}$
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% and $\Gamma \subseteq \mtypeEnvironment{}$
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% \item[given] a $(\Delta, \sigma)$ with $\Delta \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$
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% and there exists a $\Delta'$ with $\Delta, \Delta' \vdash \overline{\CC{}(\sigma(\type{S'})) <: \sigma(\type{T'})}$
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\item[then] there is a completion $|\texttt{e}|$ with $\Delta|\Gamma \vdash |\texttt{e}| : \sigma(\tv{a})$
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\end{description}
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\end{lemma}
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Regular type placeholders represent type annotations.
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These are the only types a \wildFJ{} program needs to be correctly typed.
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The type placeholders flagged as wildcard placeholders are intermediate types
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used in let statements and as type parameters for generic method calls.
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%Unify needs to return S aswell and guarantee that the \Delta' environment are the wildcards
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% only used inside the constraint the wildcard variable occurs
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% should Unify also return the \Delta' environment? Otherwise the bounds of free wildcard variables are lost
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% Or is it possible to deduct the right \ol{S} directly from the types in the normal TPHs?
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\textit{Proof:}
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By structural induction over the expression $\texttt{e}$.
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\begin{description}
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\item[$\texttt{e}.\texttt{f}$] Let $\sigma(\tv{r}) = \wcNtype{\Delta_c}{N}$,
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then $\Delta|\Gamma \vdash \texttt{e} : \wcNtype{\Delta_c}{N}$ by assumption.
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$\Delta', \Delta, \Delta_c \vdash \type{N} <: \sigma(\exptype{C}{\overline{\wtv{a}}})$ by premise.
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%Let $\sigma(\tv{r}) = \wcNtype{\Delta'}{N}$.
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%Let $\sigma([\ol{\wtv{a}}/\ol{X}]\type{T}) = \wcNtype{\Delta_t}{N_t}$.
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The completion of $|\texttt{e}.\texttt{f}|$ is $\texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f}$
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We now show
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$\Delta|\Gamma \vdash \texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f} : \sigma(a)$
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by the T-Field rule.
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$\Delta \vdash \wcNtype{\Delta_c}{N} <: \wcNtype{\Delta_c}{N}$ by S-Refl.
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$\Delta, \Delta_c \vdash \type{U}_i <: \sigma(\tv{a})$,
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because of the constraint $[\overline{\wtv{a}}/\ol{X}]\type{T} \lessdot \tv{a}$ and lemma \ref{lemma:unifySoundness}.
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$\textit{fields}(\sigma(\exptype{C}{\overline{\wtv{a}}})) = \sigma([\overline{\wtv{a}}/\ol{X}]\type{T})$
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and $\text{fv}(\type{U}_i) \subseteq \text{fv}(\type{N})$ by definition of $\textit{fields}$.
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$\text{dom}(\Delta_c) \subseteq \text{fv}{\type{N}}$ by lemma \ref{lemma:tvsNoFV}.
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% X.List<X> <. List<a?>
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% $\sigma(\ol{\tv{r}}) = \overline{\wcNtype{\Delta}{N}}$,
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% $\ol{N} <: [\ol{S}/\ol{X}]\ol{U}$,
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% TODO: S ok? We could proof $\Delta, \Delta' \overline{\Delta} \vdash \ol{S} \ \ok$
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% by proofing every substitution in Unify is ok aslong as every type in the inputs is ok
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% S ok when all variables are in the environment and every L <: U and U <: Class-bound
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% This can be given by the constraints generated. We can proof if T ok and S <: T and T <: S' then S ok and S' ok
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% If S ok and T <. S , then Unify generates a T ok
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S typeinference:
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T <: [S/Y]U
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We apply the following lemma
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Lemma
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if T ok and T <: S then S ok
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until
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T = [S/Y]U
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and then we can say by
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Lemma:
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If [S/Y]U ok then S ok (TODO: proof!)
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So we do not have to proof S ok (but T)
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% T_r <: C<T> (S is in T)
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% Is C<T> ok?
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% if every type environment \Delta supplied to Unify is ok (L <: U), then \sigma(a) = \Delta'.N implies \Delta' conforms to (L <: U)
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% this together with the X <. N constraints proofs T_r ok
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$\Delta \vdash \sigma(\tv{a}), \wcNtype{\Delta_c}{N} \ \ok$ %TODO
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%Easy, because unify only generates substitutions for normal type placeholders which are OK
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\item[$\texttt{e}.\texttt{m}(\ol{e})$]
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Lets have a look at the case where the receiver and parameter types are all named types.
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So $\sigma(\ol{\tv{r}}) = \ol{T} = \ol{\wcNtype{\Delta_u}{N}}$ and $\sigma(\tv{r}) = \type{T}_r = \wcNtype{\Delta_u}{N}$
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and a $\Delta'$, $\overline{\Delta}$ where $\Delta' \subseteq \Delta_u$, $\overline{\Delta \subseteq \Delta_u}$
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and $\text{dom}(\Delta') = (\text{fv}(\type{N})/\Delta)$, $\overline{\text{dom}(\Delta) = (\text{fv}(\type{N})/\Delta)}$ in
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$\texttt{let}\ \texttt{x} : \wcNtype{\Delta'}{N} = \texttt{e} \ \texttt{in}\
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\texttt{let}\ \ol{x} : \overline{\wcNtype{\Delta'}{N}} = \ol{e} \ \texttt{in}\ \texttt{x}.\texttt{m}(\ol{x})$
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%TODO: show that this type exists \wcNtype{\Delta'}{N} (a \Delta' which only contains free variables of the type N)
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% and which is a supertype of T_r (so no free variables in T_r)
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% Solution: Make this a lemma and guarantee that Unify does only create types Delta.T with \Delta subset fv(T)
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% This is possible because free variables (except those in \Delta_in) are never used in wildcard environments
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By lemma \ref{lemma:unifyWeakening} we know that
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$\unify{}(\Delta, [\ol{\tv{a}}/\ol{X}][\ol{\tv{b}}/\ol{Y}]\set{ \overline{\wcNtype{\Delta}{N}} \lessdotCC \ol{T}, \wcNtype{\Delta'}{N} \lessdotCC \exptype{C}{\ol{X}}})$
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has a solution.
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Also $\overline{\wcNtype{\Delta}{N}} \lessdot \ol{T}$ and $\wcNtype{\Delta'}{N} \lessdotCC \exptype{C}{\ol{X}}$ will be converted to
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$\ol{N} \lessdot \ol{T}$ and $\type{N} \lessdot \exptype{C}{\ol{X}}$ by the \rulename{Capture} rule.
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Which then results in the same as calling $\unify{}((\Delta, \Delta', \overline{\Delta}), [\ol{\tv{a}}/\ol{X}][\ol{\tv{b}}/\ol{Y}]\set{\ol{N} \lessdot \ol{T}, \type{N} \lessdot \exptype{C}{\ol{X}}})$,
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which shows $\Delta, \Delta', \overline{\Delta} \vdash \overline{N} <: [\ol{S}/\ol{X}]\overline{U}$.
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This implies $\text{dom}(\Delta') \subseteq \text{fv}(\type{N})$ and $\text{dom}(\Delta') \subseteq \text{fv}(\type{N})$
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$\Delta, \Delta', \overline{\Delta} \vdash [\ol{S}/\ol{X}]\type{U} \type{T}$,
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$\Delta, \Delta', \overline{\Delta} \vdash \ol{N} <: [\ol{S}/\ol{X}]\ol{U}$,
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and $\Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'}$
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are guaranteed by the generated constraints and lemma \ref{lemma:unifySoundness} and \ref{lemma:unifyCC}.
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$\Delta, \Delta' \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}}$ due to $\ol{T} = \ol{\wcNtype{\Delta}{N}}$ and S-Refl. $\Delta, \Delta' \vdash \type{T}_r <: \wcNtype{\Delta'}{N}$ accordingly.
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$\Delta|\Gamma \vdash \texttt{t}_r : \type{T}_r$ and $\Delta|\Gamma \vdash \ol{t} : \ol{T}_r$ by assumption.
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$\Delta, \Delta' \vdash \ol{\wcNtype{\Delta}{N}} \ \ok$ by lemma \ref{lemma:unifyWellFormedness},
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therefore $\Delta, \Delta', \overline{\Delta} \vdash \ol{N} \ \ok$, which implies $\Delta, \Delta', \overline{\Delta} \vdash \ol{U} \ \ok$
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by lemma \ref{lemma:wfHereditary} and $\Delta, \Delta', \overline{\Delta} \vdash \ol{S} \ \ok $ by premise of WF-Class.
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The same goes for $\wcNtype{\Delta'}{N}$.
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% \begin{gather}
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% \label{sp:0}
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% \Delta, \Delta', \overline{\Delta} \vdash [\ol{S}/\ol{X}]\type{U} \type{T} \\
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% \label{sp:1}
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% \Delta, \Delta', \overline{\Delta} \vdash \ol{N} <: [\ol{S}/\ol{X}]\ol{U} \\
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% \label{sp:2}
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% \Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'} \\
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% \label{sp:3}
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% \Delta, \Delta', \overline{\Delta} \vdash \ol{S} \ \ok \\
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% \label{sp:4}
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% \Delta, \Delta' \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}}
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% \end{gather}
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Method calls generate multiple constraints that share the same wildcard placeholders ($\ol{\wtv{a}}$, $\ol{\wtv{b}}$).
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\end{description}
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% \begin{lemma} Unify does add free variables to types not containing free variables (or wildcard placeholders)
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% \begin{description}
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% \item[If] $(\sigma, \Delta) = \unify{}( \Delta', C )$
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% \item[Then] $\forall x \in \set{\type{S} \mid \type{S} \in C, \text{fv}(\type{S}) = \emptyset }: \text{fv}(\sigma(\type{S})) \subseteq \text{dom}(\Delta)$
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% \end{description}
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% \end{lemma}
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% \textit{Proof:} by induction over the \unify{} algorithm.
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% A unifier $\sigma(\tv{a}) = \type{T}$, where the type variable $\tv{a}$ is not flagged as a wildcard will always hold
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% $\text{fv}(\type{T}) \subseteq \text{dom}(\Delta)$.
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% %UNIFY fails when there are free variables on the right side of a a =. T constraint
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\begin{lemma}\label{lemma:unifyNoExcessWildcards}
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\begin{description}
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\item[If] $(\Delta, \sigma) = \unify{}(\Delta', C)$
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\item[Then] $\sigma(\tv{a}) = \wcNtype{\Delta''}{N} \implies \text{dom}(\Delta'') \subseteq \text{fv}(\type{N})$
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\end{description}
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\end{lemma}
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\textit{Proof:} Easy -
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All types given in the input to the \unify{} algorithm already comply with this requirement
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and none of the rules change.
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Note: the \rulename{Super} rule removes unnecessary wildcards.
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\begin{lemma}\label{lemma:unifyWellFormedness}
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\unify{} generates well-formed types as long as well-formed types are supplied.
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\begin{description}
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\item[If] $(\Delta, \sigma) = \unify{}(\Delta', C)$ %and $\sigma(\tv{a}) = \wcNtype{\ol{\wildcard{X}{\type{U}}{\type{L}}}}{N}$
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\item[and] $\wcNtype{\Delta}{N} \in C \implies \Delta' \vdash \wcNtype{\Delta}{N} \ \ok$
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%\item[then] $\Delta \vdash \ol{L <: U}$ and $\Delta \vdash \ol{U <: U'}$ % (with U' being upper limit by class definition)
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\item[then] $\Delta \vdash \sigma(\tv{a}) \ \ok$
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\end{description}
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\end{lemma}
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%Only the \rulename{General} rule generates fresh wildcards.
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%By lemma \ref{lemma:unifySoundness} we get $\Delta \vdash \sigma(\type{T}) <: \sigma(\tv{a})$ with $\sigma(\tv{a}) = \wctype{\ol{\wildcard{X}{\sigma(U)}{\sigma(L)}}}{C}{\ol{X}}$
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%By S-Exists and S-Trans we can say $\Delta \vdash \sigma(\type{L}) <: \sigma(\type{U})$
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\textit{Proof:}
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by induction over every step of the \unify{} algorithm.
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Hint: a type placeholder $\tv{a}$ will never be replaced by a free variable or a type containing free variables.
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This fact together with the presumption that every supplied type is well-formed we can easily show that this lemma is true.
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\textit{Proof: (Variant 2)}
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The GenSigma and GenDelta rules check for well-formedness.
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% All types supplied to the Unify algorithm by TYPEs only contain ? extends or ? super wildcards. (X : [bot..T] or X : [T .. Object]).
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% Both are well formed!
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\begin{lemma}\label{lemma:wfHereditary}
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Well-formedness is hereditary
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\begin{description}
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\item[If] $\triangle \vdash \exptype{C}{\ol{X}} \ \ok$ and $\triangle \vdash \exptype{C}{\ol{X}} <: \type{S}$
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\item[Then] $\triangle \vdash \type{S} \ \ok$
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\end{description}
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\end{lemma}
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\textit{Proof:}
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\begin{lemma}
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\begin{description}
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\item[If] $\Delta \vdash \wctype{\Delta'}{C}{\ol{T}} \ \ok$
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\item[Then] $\Delta, \Delta' \vdash \ok{T} \ \ok$
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\end{description}
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\end{lemma}
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\textit{Proof:} by definition of WF-Class
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\begin{lemma} \label{lemma:tvsNoFV}
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\unify{} does not add free variables to types not containing free variables.
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\begin{description}
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\item[If] $(\sigma, \Delta) = \unify{} (\Delta', C)$
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\item[and] $\tv{a}$ being a type placeholders used in $C$
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\item[then] $\text{fv}(\sigma(\tv{a})) \subseteq \Delta, \Delta'$
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\end{description}
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\end{lemma}
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\textit{Proof:}
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Trivial. \unify{} fails when a constraint $\tv{a} \doteq \rwildcard{X}$ arises.
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% do we even have to proof that?
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% \begin{lemma}
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% C = Delta.N <c T
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% is the same for Unify as:
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% Delta |- N <. T
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% \end{lemma}
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\begin{lemma} \label{lemma:unifyWeakening}
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Removing constraints does not render \unify{} impossible as long as the removed constraints do not share wildcard placeholders.
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If there is a solution for a constraint set $C$, then there is also a solution for a subset of that constraint set.
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\begin{description}
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\item[If] $(\sigma, \Delta) = \unify{}( C \cup C')$ and $\text{wtv}(C) \cap \text{wtv}(C') = \emptyset$
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\item[then] $ (\sigma', \Delta') = \unify{}( C')$
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\end{description}
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\end{lemma}
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\textit{Proof:}
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%TODO
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% does this really work. We have to show that the algorithm never gets stuck as long as there is a solution
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% maybe there are substitutions with types containing free variables that make additional solutions possible. Check!
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\begin{lemma}{\unify{} Soundness:}\label{lemma:unifySoundness}
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\unify{}'s type solutions are correct respective to the subtyping rules defined in figure \ref{fig:subtyping}.
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\begin{description}
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\item[If] $(\sigma, \Delta) = \unify{}( \Delta', \, \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \type{S'} \lessdotCC \type{T'} } )$ %\cup \overline{ \type{S} \doteq \type{S'} })$
|
|
\item[Then] there exists a substitution $\sigma'$ and a set of types $\overline{\wcNtype{\Delta}{N}}$ with:
|
|
\begin{itemize}
|
|
\item $\sigma \subseteq \sigma'$
|
|
\item $\Delta, \Delta' \vdash \overline{\sigma'(\type{S}) <: \sigma'(\type{T})}$
|
|
\item $\Delta, \Delta' \vdash \overline{\sigma'(\type{S'}) <: \wcNtype{\Delta}{N}}$
|
|
\item $\Delta, \Delta', \overline{\Delta} \vdash \overline{\type{N} <: \sigma'(\type{T'})}$
|
|
\end{itemize}
|
|
\end{description}
|
|
\end{lemma}
|
|
|
|
% \begin{lemma} % a lemma where we distinguis between free variable on the left or the right side of a constraint (not needed anymore)
|
|
% The \unify{} algorithm only produces correct output for constraints not containing free variables.
|
|
% \begin{description}
|
|
% \item[If] $(\sigma, \Delta) = \unify{}( \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \wcNtype{\Delta'}{N} \lessdot \type{T'} } \cup \overline{ \type{S'} \lessdot \wcNtype{\Delta'}{N'} })$
|
|
% \item[and] $fv(\overline{ \type{S} }) = \emptyset, fv(\overline{ \type{T} }) = \emptyset, fv(\overline{ \type{T'} }) = \emptyset, fv(\overline{ \type{S'} }) = \emptyset$
|
|
% \item[Then] $\Delta \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$
|
|
% and $\Delta, \Delta' \vdash \overline{\sigma(\type{N}) <: \sigma(\type{T'})}, \overline{\sigma(\type{S'}) <: \sigma(\type{N'})}$
|
|
% %TODO: Rephrase (\Delta' is used three times!)
|
|
% %The function $\CC{}$ is given as $\CC{}(\wcNtype{\Delta}{N}) = \type{N}$
|
|
% \end{description}
|
|
% \end{lemma}
|
|
\textit{Proof:}
|
|
%(we are going backwards over the algorithm)
|
|
%first we have to determine the \Delta'' -> it's only the wildcards which are free in N
|
|
% during this proof we can use Delta'' as we like
|
|
|
|
For every step in the \unify{} algorithm:
|
|
Assuming the unifier $\sigma$ is correct for a constraint set $C'$, the unifier is also correct for the
|
|
constraint set $C$ before the transformation.
|
|
% \begin{description}
|
|
% \item[Assumption:] $\unify{}(C) = (\Delta, \sigma)$, with $\Delta \vdash \sigma(C)$
|
|
% \item[Induction step:] For every case $C'$ which can be transformed to $C$ we have to show $\Delta \vdash \sigma(C')$
|
|
% \end{description}
|
|
|
|
\unify{} terminates with $C = \emptyset$ for which the preposition holds:
|
|
$\Delta \vdash \sigma(\emptyset)$
|
|
|
|
We now show that for every transformation of a constraint set $C$ to a constraint set $C'$
|
|
the preposition holds for $C$ using the assumption that it holds for $C'$ :
|
|
$\Delta \vdash \sigma(C') \implies \Delta \vdash \sigma(C)$
|
|
|
|
\begin{description}
|
|
\item[AddDelta] $C$ is not changed
|
|
\item[GenDelta] by definition, S-Var-Left, and S-Trans %The generated type variable is unique due to the solved form property. %and the solved form property (no $\tv{a}$ in $C$)
|
|
\item[GenDelta'] same as GenDelta by setting $\sigma'(\wtv{b}) = \rwildcard{B}$
|
|
\item[GenSigma] by definition and S-Refl.
|
|
% holds for $\set{\tv{a} \doteq \type{G}}$ by definition.
|
|
% Holds for $C$ by assumption and because $\tv{a} \notin C$ by solved form definition ($[\type{G}/\tv{a}]C = C$).
|
|
\item[Ground] Assumption and S-Bot
|
|
\item[Sub-Elim] Assumption and S-Refl
|
|
\item[Force] by assumption and $\rwildcard{X} = \type{U}$ %TODO: step 5 should remove all X^T_T with T (make wildcards with same upper and lower bounds to normal types)
|
|
\item[Raise] Assumption, S-Trans
|
|
\item[Settle] Assumption, S-Trans
|
|
\item[Super] S-Extends ($\vdash \wctype{\Delta}{C}{\ol{T}} <: \wctype{\Delta}{D}{[\ol{T}/\ol{X}]\ol{N}}$), S-Trans
|
|
\item[\generalizeRule{}] by Assumption, because $C \subset C'$
|
|
\item[Same] by S-Exists
|
|
\item[SameW] %TODO
|
|
\item[Adapt] Assumption, S-Extends, S-Trans
|
|
\item[Adopt] Assumption, because $C \subset C'$
|
|
%\item[Capture, Reduce] are always applied together. We have to destinct between two cases:
|
|
\item[Prepare]
|
|
Given a $\wctype{\Delta_c}{C}{\ol{S}} \lessdot \wcNtype{\Delta''}{N}$
|
|
we get $\Delta, \Delta', \overline{\Delta} \vdash \exptype{C}{\ol{S}} <: \wcNtype{\Delta''}{N}$ with $\Delta_c \subseteq \overline{\Delta}$.
|
|
|
|
$\text{fv}(\sigma'(\wctype{\Delta_c}{C}{\ol{S}})) = \emptyset$ implies $\text{fv}(\ol{S})\subseteq \text{dom}(\Delta, \Delta_c)$
|
|
and $\text{fv}(\sigma'(\wcNtype{\Delta''}{N})) = \emptyset$ implies $\text{dom}(\Delta_c) \cap \text{fv}(\sigma'(\wcNtype{\Delta''}{N})) = \emptyset$.
|
|
|
|
No free variables on both sides also mean we do not need $\overline{\Delta}$.
|
|
Therefore we can say $\Delta, \Delta' \vdash \wctype{\Delta_c}{C}{\ol{S}} <: \wcNtype{\Delta''}{N}$.
|
|
|
|
\item[Capture]
|
|
Everytime the \rulename{Capture} rule is invoked we add the freshly generated free variables to the global environment $\wildcardEnv$.
|
|
We get a $\sigma'$ %and a $\Delta'$
|
|
with $\Delta, \Delta' \vdash \sigma'([\ol{\rwildcard{C}}/\ol{\rwildcard{B}}] \exptype{C}{\ol{S}}) <: \sigma'(\wctype{\Delta}{C}{\ol{T}})$
|
|
where $\sigma'(\ol{\wildcard{C}{U}{L}}) \subseteq \Delta'$ by assumption.
|
|
\unify{} performs a capture conversion only on $\lessdotCC$ constraints.
|
|
Therefore we can say that $\Delta, \Delta', \overline{\Delta} \vdash \sigma'(\exptype{C}{\ol{S}})
|
|
<: \sigma'(\wctype{\Delta}{C}{\ol{T}})$ with $\overline{\Delta}$ being all the fresh wildcards generated by \rulename{Capture}.
|
|
|
|
\item[Reduce]
|
|
|
|
%Assumption and S-Exists.
|
|
% Three different cases of the constraint $\exptype{C}{\ol{S}} \lessdot \wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}$:
|
|
% \begin{description}
|
|
% \item[$\text{fv}(\exptype{C}{\ol{S}}) = \emptyset, \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}})$:]
|
|
% the preposition holds by Assumption and S-Exists.
|
|
% \item[$\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset$:]
|
|
% then $\text{fv}(\exptype{C}{\ol{S}}) \subseteq \Delta'$ with $\Delta' = \overline{\wildcard{A}{\type{U}}{\type{L}}}$
|
|
% \item[$\text{fv}(\exptype{C}{\ol{S}}) = \emptyset$] $\Delta' = \emptyset$
|
|
% \end{description}
|
|
% List<X> <. Y.List<Y>, free variables are either in
|
|
If $\text{fv}(\exptype{C}{\ol{S}}) = \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset$ the preposition holds by Assumption and S-Exists.
|
|
Otherwise
|
|
$\Delta' \vdash \CC{}(\sigma(\exptype{C}{\ol{S}})) <: \sigma(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}})$
|
|
holds with any $\Delta'$ so that $(\text{fv}(\exptype{C}{\ol{S}}) \cup \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) ) \subseteq \text{dom}(\Delta') $.
|
|
\item[Match] Assumption, S-Trans
|
|
\item[Trim] Assumption and S-Exists
|
|
\item[Remove] $C$ is not changed
|
|
\item[Circle] S-Refl
|
|
\item[Swap] by definition
|
|
\item[Erase] S-Refl
|
|
\item[Equals] by definition \ref{def:equal}
|
|
%by definition
|
|
%TODO
|
|
% Unify does never contain wildcard environments with unused wildcards. Therefore after N <: N' and N' <: N, both types have the same wildcard environment
|
|
|
|
|
|
% \item[Reduce]
|
|
% The renaming from $\rwildcard{C}$ to $\rwildcard{B}$ is not a problem. It's allowed to rename wildcards inside a type.
|
|
% Removing $\rwildcard{C}$ from the environment does not change anything because it was freshly generated and is used nowhere.
|
|
|
|
% The rest follows directly from S-Exists.
|
|
% We can say: $\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset$,
|
|
% because the input to the \unify{} algorithm has no free type variables and we never substitute a type with free type variables
|
|
% and none of the other steps of the algorithm generates a $\lessdot$ constraint containing free type variables on the right side. %TODO: proof
|
|
|
|
% $\text{fv}(\ol{T}) \subseteq \text{dom}(\wildcardEnv \cup \overline{\wildcard{B}{\type{U'}}{\type{L'}}})$
|
|
% TODO: The capture conversion has to be when substituting a $\wtv{a}$ variable. Then we have to rename!
|
|
% %Lets first try it without the capture conversion. And involve the wtvs in the second step
|
|
% % The algorithm works by never substituting wildcards
|
|
|
|
|
|
|
|
% \unify{} cannot guarantee the premise $\text{dom}(\Delta, \Delta') \cap \text{fv}(\wcNtype{\overline{\wildcard{X}{\type{U}}{\type{L}}}}{N}) = \emptyset$.
|
|
% We loosen the soundness requirements and allow a arbitrary environment $\Delta''$ to be added to the right side of the subtype relation.
|
|
% This is still sufficient to proof soundness for the whole algorithm.
|
|
% We show that the need for the additional environment $\Delta''$ can be satisfied by a let statement.
|
|
|
|
% \begin{itemize}
|
|
% \item $\Delta, \Delta' \vdash [\ol{T}/\ol{X}]\ol{L} <: \ol{T}$: S-Exists
|
|
% \item $\Delta, \Delta' \vdash \ol{T} <: [\ol{T}/\ol{X}]\ol{U}$: S-Exists
|
|
% \item $\textit{fv}(\ol{T}) \subseteq \text{dom}{\Delta, \Delta'}$: $\wildcardEnv$ holds all variables %TODO: Proof
|
|
% \end{itemize}
|
|
\item[Normalize] Assumption and lemma 5 \emph{substitution preserves subtyping}.%\ref{lemma:wildcardReplacement}. (Or Lemma 5 from the wildcard paper. \emph{substitution preserves subtyping})
|
|
% The GenSigma step replaces both sides of $\rwildcard{A} \doteq \rwildcard{B}$ with the upper bound $\type{U}$.
|
|
% This works for every constraint, whether it contains free variables or not.
|
|
% It does not add to free variables of constraints because the upper bound does not contain any.
|
|
The GenSigma and Gen Delta steps remove Wildcards which have the same upper and lower bound.
|
|
$\rwildcard{A},\rwildcard{B} \notin \sigma(C)$
|
|
|
|
% sigma(T) = sigma(U) we have to show that T = U means \Delta \vdash [T/U]C \implies \Delta \vdash [U/T]C
|
|
% the constraints L <. U, U <. L lead to L =. U
|
|
%If L is List<X> with X being free wildcard
|
|
%then U <. L will fail if U is type variable
|
|
|
|
|
|
|
|
|
|
% this is because bounds never contain free variables (is that true?)
|
|
|
|
%This type contains free variables when A is replaced by an CC wildcard
|
|
|
|
%This must fail:
|
|
\begin{verbatim}
|
|
<A> A m(List<? extends List<A>> l, A)
|
|
|
|
m(List<List<? super String>> l, "hi")
|
|
\end{verbatim}
|
|
%This fails because of Equals rule (TODO: proof)
|
|
|
|
\item[Tame] same reasoning as Normalize
|
|
\item[Bot] S-Bot
|
|
\item[Pit] S-Bot
|
|
\item[Upper] S-Trans and S-VarLeft
|
|
\item[Lower] S-Trans and S-VerRight
|
|
\item[Subst-WC] by S-Refl
|
|
\item[Subst]
|
|
$\sigma(C \cup \set{\tv{a} \doteq \type{T}}) = \sigma([\type{T}/\tv{a}]C \cup \set{\tv{a} \doteq \type{T}})$
|
|
and
|
|
$\sigma(\wildcardEnv) = \sigma([\type{T}/\tv{a}]\wildcardEnv)$
|
|
\item[Subst-WC]
|
|
%Proof by Lemma 5 \emph{Type substitution preserves subtyping} from \cite{WildcardsNeedWitnessProtection}.
|
|
Same as Subst
|
|
\end{description}
|
|
|
|
|
|
\subsection{Converting to Wild FJ}
|
|
Wildcards are existential types which have to be \textit{unpacked} before they can be used.
|
|
In Java this is done implicitly by a process called capture conversion \cite{JavaLanguageSpecification}.
|
|
The type system in \cite{WildcardsNeedWitnessProtection} makes this process explicit by using \texttt{let} statements.
|
|
Our type inference algorithm will accept an input program without let statements and add them where necessary.
|
|
|
|
%Type inference adds \texttt{let} statements in a fashion similar to the Java capture conversion \cite{WildFJ}.
|
|
%We wrap every parameter of a method invocation in \texttt{let} statement unknowing if a capture conversion is necessary.
|
|
|
|
Figure \ref{fig:tletexpr} shows type rules for fields and method calls.
|
|
They have been merged with let statements and simplified.
|
|
|
|
The let statements and the type $\wcNtype{\Delta'}{N}$, which the type inference algorithm can freely choose,
|
|
are necessary for the soundness proof.
|
|
|
|
%TODO: Show that well-formed implies witnessed!
|
|
We change the type rules to require the well-formedness instead of the witnessed property.
|
|
See figure \ref{fig:well-formedness}.
|
|
|
|
Our well-formedness criteria is more restrictive than the ones used for \wildFJ{}.
|
|
\cite{WildcardsNeedWitnessProtection} works with the two different judgements $\ok$ and \texttt{witnessed}.
|
|
With \texttt{witnessed} being the stronger one.
|
|
We rephrased the $\ok$ judgement to include \texttt{witnessed} aswell.
|
|
$\type{T} \ \ok$ in this paper means $\type{T} \ \ok$ and $\type{T} \ \texttt{witnessed}$ in the sense of the \wildFJ{} type rules.
|
|
Java's type system is complex enough as it is. Simplification, when possible, is always appreciated.
|
|
Our $\ok$ rule may not be able to express every corner of the Java type system, but is sufficient to show soundness regarding type inference for a Java core calculus.
|
|
|
|
The \rulename{WF-Class} rule requires upper and lower bounds to be direct subtypes of each other $\type{L} <: \type{U}$.
|
|
The type rules from \cite{WildcardsNeedWitnessProtection} use a witness type instead.
|
|
Stating that a type is well formed (\texttt{witnessed}) if there exists atleast one simple type $\type{N}$ as a possible subtype:
|
|
$\Delta \vdash \type{N} <: \wctype{\Delta'}{C}{\ol{T}}$.
|
|
|
|
A witness type is easy to find by replacing every wildcard $\ol{W}$ in $\wctype{\ol{\wildcard{W}{U}{L}}}{C}{\ol{T}}$ by its upper bound:
|
|
$\Delta \vdash \exptype{C}{[\ol{U}/\ol{W}]\ol{T}} <: \wctype{\ol{\wildcard{W}{U}{L}}}{C}{\ol{T}}$.
|
|
$\Delta \vdash \exptype{C}{[\ol{U}/\ol{W}]\ol{T}} \ \texttt{witnessed}$ is given due to $\Delta \vdash \ol{T}, \ol{L}, \ol{U} \ \ok$
|
|
and $\Delta \vdash \wctype{\ol{\wildcard{W}{U}{L}}}{C}{\ol{T}} \ \ok$.
|
|
|
|
\begin{figure}[tp]
|
|
$\begin{array}{l}
|
|
\typerule{T-Var}\\
|
|
\begin{array}{@{}c}
|
|
x : \type{T} \in \Gamma
|
|
\\
|
|
\hline
|
|
\vspace*{-0.3cm}\\
|
|
\Delta | \Gamma \vdash x : \type{T}
|
|
\end{array}
|
|
\end{array}$
|
|
\\[1em]
|
|
$\begin{array}{l}
|
|
\typerule{T-New}\\
|
|
\begin{array}{@{}c}
|
|
\Delta, \overline{\Delta} \vdash \exptype{C}{\ol{T}} \ \ok \quad \quad
|
|
\text{fields}(\exptype{C}{\ol{T}}) = \overline{\type{U}\ f} \quad \quad
|
|
\Delta | \Gamma \vdash \overline{t : \type{S}} \quad \quad
|
|
\Delta \vdash \overline{\type{S}} <: \overline{\wcNtype{\Delta}{N}} \\
|
|
\Delta, \overline{\Delta} \vdash \overline{\type{N}} <: \overline{\type{U}} \quad \quad
|
|
\Delta, \overline{\Delta} \vdash \exptype{C}{\ol{T}} <: \type{T} \quad \quad
|
|
\overline{\text{dom}(\Delta) \subseteq \text{fv}(\type{N})} \quad \quad
|
|
\Delta \vdash \type{T}, \overline{\wcNtype{\Delta}{N}} \ \ok
|
|
\\
|
|
\hline
|
|
\vspace*{-0.3cm}\\
|
|
\Delta | \Gamma \vdash \letstmt{\ol{x} : \ol{\wcNtype{\Delta}{N}} = \ol{t}}{\texttt{new} \ \exptype{C}{\ol{T}}(\overline{t})} : \type{T}
|
|
\end{array}
|
|
\end{array}$
|
|
\\[1em]
|
|
$\begin{array}{l}
|
|
\typerule{T-Field}\\
|
|
\begin{array}{@{}c}
|
|
\Delta | \Gamma \vdash \texttt{t} : \type{T} \quad \quad
|
|
\Delta \vdash \type{T} <: \wcNtype{\Delta'}{N} \quad \quad
|
|
\textit{fields}(\type{N}) = \ol{U\ f} \\
|
|
\Delta, \Delta' \vdash \type{U}_i <: \type{S} \quad \quad
|
|
\text{dom}(\Delta') \subseteq \text{fv}(\type{N}) \quad \quad
|
|
\Delta \vdash \type{S}, \wcNtype{\Delta'}{N} \ \ok
|
|
\\
|
|
\hline
|
|
\vspace*{-0.3cm}\\
|
|
\Delta | \Gamma \vdash \texttt{let}\ x : \wcNtype{\Delta'}{N} = \texttt{t} \ \texttt{in}\ x.\texttt{f}_i : \type{S}
|
|
\end{array}
|
|
\end{array}$
|
|
\\[1em]
|
|
$\begin{array}{l}
|
|
\typerule{T-Call}\\
|
|
\begin{array}{@{}c}
|
|
\Delta, \Delta', \overline{\Delta} \vdash \ol{\type{N}} <: [\ol{S}/\ol{X}]\ol{U} \quad \quad
|
|
\textit{mtype}(\texttt{m}, \type{N}) = \generics{\ol{X \triangleleft U'}} \ol{U} \to \type{U} \quad \quad
|
|
\Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'}
|
|
\\
|
|
\Delta, \Delta', \overline{\Delta} \vdash \ol{S} \ \ok \quad \quad
|
|
\Delta | \Gamma \vdash \texttt{t}_r : \type{T}_r \quad \quad
|
|
\Delta | \Gamma \vdash \ol{t} : \ol{T} \quad \quad
|
|
\Delta \vdash \type{T}_r <: \wcNtype{\Delta'}{N} \quad \quad
|
|
\Delta \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}}
|
|
\\
|
|
\Delta \vdash \type{T}, \wcNtype{\Delta'}{N}, \overline{\wcNtype{\Delta}{N}} \ \ok \quad \quad
|
|
\Delta, \Delta', \Delta'' \vdash [\ol{S}/\ol{X}]\type{U} <: \type{T} \quad \quad
|
|
\text{dom}(\Delta') \subseteq \text{fv}(\type{N}) \quad \quad
|
|
\overline{\text{dom}(\Delta) \subseteq \text{fv}(\type{N})}
|
|
\\
|
|
\hline
|
|
\vspace*{-0.3cm}\\
|
|
\Delta | \Gamma \vdash \letstmt{x : \wcNtype{\Delta'}{N} = t_r, \ol{x} : \ol{\wcNtype{\Delta}{N}} = \ol{t}}
|
|
{\texttt{x}.\generics{\ol{S}}\texttt{m}(\ol{x})} : \type{T}
|
|
\end{array}
|
|
\end{array}$
|
|
\\[1em]
|
|
$\begin{array}{l}
|
|
\typerule{T-Elvis}\\
|
|
\begin{array}{@{}c}
|
|
\Delta | \Gamma \vdash t_1 : \type{T}_1 \quad \quad
|
|
\Delta | \Gamma \vdash t_2 : \type{T}_2 \quad \quad
|
|
\Delta \vdash \type{T}_1 <: \type{T} \quad \quad
|
|
\Delta \vdash \type{T}_2 <: \type{T}
|
|
\\
|
|
\hline
|
|
\vspace*{-0.3cm}\\
|
|
\Delta | \Gamma \vdash t_1 \elvis{} t_2 : \type{T}
|
|
\end{array}
|
|
\end{array}$
|
|
\\[1em]
|
|
$\begin{array}{l}
|
|
\typerule{T-Method}\\
|
|
\begin{array}{@{}c}
|
|
\text{dom}(\Delta)=\ol{X} \quad \quad
|
|
\Delta' = \overline{\type{Y} : \bot .. \type{U}} \quad \quad
|
|
\Delta, \Delta' \vdash \ol{U}, \type{T}, \ol{T}\ \ok \quad \quad
|
|
\texttt{class}\ \exptype{C}{\ol{X \triangleleft \_ }} \triangleleft \type{N} \set{\ldots} \\
|
|
\Delta, \Delta' | \overline{x:\type{T}}, \texttt{this} : \exptype{C}{\ol{X}} \vdash t:\type{S} \quad \quad
|
|
\Delta, \Delta' \vdash \type{S} <: \type{T} \quad \quad
|
|
\text{override}(\texttt{m}, \type{N}, \generics{\ol{Y \triangleleft U}}\ol{T} \to \type{T})
|
|
\\
|
|
\hline
|
|
\vspace*{-0.3cm}\\
|
|
\Delta \vdash \generics{\ol{Y \triangleleft U}} \type{T} \ \texttt{m}(\overline{\type{T} \ x}) = t \ \ok \ \texttt{in C}
|
|
\end{array}
|
|
\end{array}$
|
|
\\[1em]
|
|
$\begin{array}{l}
|
|
\typerule{T-Class}\\
|
|
\begin{array}{@{}c}
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\Delta = \overline{\type{X} : \bot .. \type{U}} \quad \quad
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\Delta \vdash \ol{U}, \ol{T} \ \ok \quad \quad
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\Delta \vdash \type{N} \ \ok \quad \quad
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\Delta \vdash \ol{M} \ \ok \texttt{ in C}
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\\
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\hline
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\vspace*{-0.3cm}\\
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\texttt{class}\ \exptype{C}{\ol{X \triangleleft U}} \triangleleft \type{N} \set{\overline{\type{T}\ f}; \ol{M} } \ \ok
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\end{array}
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|
\end{array}$
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\caption{T-Call and T-Field} \label{fig:tletexpr}
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\end{figure}
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|
|